A 39-year annuity-immediate will pay 13 in each of the first 3 years, 12 in each of the next 3 years, etc., until payments of 1 are made in each of the last 3 years. The present value of the payments at an annual effective rate of 3% is $X$. Determine $X$.
There are a few ways to do this, but this is my attempt.
$X=13v +13v^2+13v^3+12v^4+12v^5+...+v^{39}$
$X=13(v+v^2+v^3)+12v^3(v+v^2+v^3)+v^{36}(v+v^2+v^3)$
$X=(v+v^2+v^3)(13+12v^3+11v^6+...+v^{36})$
$X=a_{3^\urcorner.03}(Da)_{13^\urcorner j}=(2.828)(60.617)=171.425$
Where $j=(1.03)^3-1=.092727$ is the three year effective rate. This strategy seems straightforward but it's wrong. The book its from offers a similar solution: $X=s_{3^\urcorner.03}(Da)_{13^\urcorner j}=187.32$, i.e. our answers differ by a factor of $1.03^3$. Where did I go wrong?
Let $i$ be the interest rate. $$X=\left ( \frac{1}{1+i}+\frac{1}{(1+i)^2}+\frac{1}{(1+i)^3} \right )\sum_{k=1}^{13}\frac{k}{(1+i)^{39-3k}} $$ The sum is in the form $$ \sum_{k=1}^{n}kz^k=z\frac{1-(n+1)z^n+nz^{n+1}}{(1-z)^n} $$ so $$ \sum_{k=1}^{13}\frac{k}{(1+i)^{39-3k}}=\frac{1}{(1+i)^{39}}\sum_{k=1}^{13}k(1+i)^{3k}=\frac{1}{(1+i)^{39}}(1+i)^{3}\frac{1-14(1+i)^{39}+13(1+i)^{42}}{(1-(1+i)^3)^2}=66.238 $$ Multiplying this by the other term gives $X=187.362$